Question

\(\dfrac{x+2\sqrt{x}+1}{x-1}\)+\(\dfrac{x-\sqrt{x}}{x-2\sqrt{x}+1}\)-\(\dfrac{x-2\sqrt{x}}{x-\sqrt{x}}\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)

\(\dfrac{x+2\sqrt{x}+1}{x-1}+\dfrac{x-\sqrt{x}}{x-2\sqrt{x}+1}-\dfrac{x-2\sqrt{x}}{x-\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}-\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}+1+\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\)